Talk:Gravity of Earth/Archive 1

Pluto

Should Pluto be removed from the list of planetary gravities? Asteron 21:22, 14 September 2006 (UTC)

Centrifugal Force Elimination Plague

This article has joined the legion of other Wikipedia articles being religiously edited for removal of centrifugal force. This would be ok with me if someone came up with a clear and simple way of explaining a given phenomenon without a rotating frame of reference, but I have yet to see an example where this is done - in fact, as far as I can tell I haven't seen an example where someone replaces with an explanation that is accurate, let alone done in a way that can be understood easily. In point of fact, someone has listened too closely a to zealous science teacher who has disregarded the model of a rotating frame of reference and decided that centrifugal force doesn't exist. Rotating frames of reference are useful, and we all understand them quite intuitively since we realized as children that it was possible to run around a corner or spin in circles, therefore centrifugal force is useful and should not be removed unless it is replaced with a superior and simple way to explain the same thing. If nobody responds to this talk within a day or so I'm going to edit centrifugal force back into this article. (Yes, I borrowed some of this rant from an xkcd comic, but I paraphrased.) --Cesoid (talk) 02:09, 24 July 2009 (UTC)

I tend to agree. Centripetal force is whatever force exists to maintain a circular motion. In the case of the Earth, centripetal force is exactly equivalent to gravitational force. Centrifugal force is a useful concept describe the experienced outward counteracting tendency caused by rotation, as experienced by an observer in that rotating frame. Yes, we know that it isn't a real force as seen by an observer in a distant inertial frame of reference. But to simply replace the word "centrifugal" with "centripetal" does far more damage to understanding the concept of why the Earth bulges out at the equator. To be more specific, at the Earth's equator, centripetal force (gravity) is about −9.8 m/s^2, but centrifugal force is about +0.04 m/s^2. Confuse one with the other, and you are off by two orders of magnitude! CosineKitty (talk) 15:52, 6 November 2009 (UTC)

Follow-up: I have corrected what I believe to be the worst of the confusion between centripetal and centrifugal in this article. Both concepts need to be here, and I hope my explanations are persuasive. If not, please discuss here so we can come to an agreement about how to phrase things to be as clear and accurate as possible to both seasoned physicists and to people new to the subject. CosineKitty (talk) 16:08, 6 November 2009 (UTC)

Regarding the sentence "At the poles the radius of curvature is zero, so only this weakened gravity would contribute to weight and objects would not float.": I'm pretty sure that the radius of curvature at the poles is not zero, and even if it were, that wouldn't be the reason why there's no "centrifugal force" there.Philgp (talk) 18:51, 3 April 2010 (UTC)

That is a good point. I have revised the wording to be more clear. There were two problems: the use of is instead of would for a hypothetical case of weakened gravity, and use of an ambiguous phrase radius of curvature. I think I know what the original author meant: that an object sitting on the surface of the Earth at one of the poles can be thought of as moving around in a circle with a radius of zero due to the Earth's rotation. The problem is, radius of curvature could mean the curvature of the Earth's surface at the poles, which would not be zero in any case. CosineKitty (talk) 19:44, 4 April 2010 (UTC)

Earth's gravitational field strength

Does anyone know the Earth's gravitational field strength (N/kg)in 3 decimal points? (Anonymous comment by User:213.105.217.232)

1. Definition of a Newton: $1\,\mathrm {N} =1\,{\frac {\mathrm {kg} \cdot \mathrm {m} }{\mathrm {s} ^{2}}}$
2. Divide both sides by 1 kg: $1\,{\frac {\mathrm {N} }{\mathrm {kg} }}=1\,{\frac {\mathrm {kg} \cdot \mathrm {m} }{\mathrm {s} ^{2}\cdot \mathrm {kg} }}$
3. Cancel the kgs on the right side: $1\,{\frac {\mathrm {N} }{\mathrm {kg} }}=1\,{\frac {\mathrm {m} }{\mathrm {s} ^{2}}}$

Thus N/kg is the same as m/s^2, so any value in m/s^2 in the article can be used. Standard acceleration, rounded to three decimal places, would then be 9.807 N/kg. —Ben Brockert (42) UE News 07:14, 6 May 2007 (UTC)

Gravitational field of Earth from Nasa

Someone add the picture and information from this article please.--Svetovid 20:08, 2 June 2007 (UTC)

Better pictures + complete info about the project: [1]--Svetovid 20:10, 2 June 2007 (UTC)

If anyone could add some sort of reference, that'd be great. For example, something simple like state that it varies by 3 percent from weakest to strongest areas. Or have an actual measure of strength in some places.

Anonymous 3:22, 11 March 2012 (UCT) — Preceding unsigned comment added by 96.51.61.155 (talk)

Table of comparative gravities: Centrifugally-adjusted value of g is incorrect (I think)

There is a lot of precision in the section Comparative gravities of the Earth, Sun, Moon and planets. First, the text above the table states that the values are for equatorial surface gravity. However, the table normalizes Earth to a value of “1” and further states that this is equivalent to an acceleration of 9.806

65 m/s². It seems to me that something isn’t quite right and—at this level of precision—we’re comparing apples and oranges. The value 9.80665 m/s² takes into account the centrifugal force of Earth’s rotation. The value 9.80665 m/s² corresponds to a perfect-sphere, sea-level latitude of 50.53°. Not surprisingly, this latitude isn’t too different from that of Sèvres (Paris, which is 49° at an altitude 50 m where the BIPM is located). For many planets on Wikipedia, the surface gravity is at the equator. For still others, like Pluto, “equatorial gravity” isn’t known and it is left out. Are we to use equatorial gravity for Earth? If so, then the value certainly would not be 9.80665 m/s². It is perfectly appropriate for Wikipedia articles on individual planets to mention equatorial gravity (where available). But for a comparison table, where this sort of data isn’t always available for every planet, the values for gravity should be calculated based solely on two criteria: mass, and mean radius. If one does the same for Earth, then its mass (5.9736 × 10²⁴ kg per Wikipedia), and volumetric radius (6,371,000.8 m), and the value of G from Wikipeida (6.67428(67) × 10^–11 m³kg^–1s^–2), means that simple g = 9.8226 m/s².

At what speed will someone situated over Jupiter’s equator at a pressure altitude of 75 kPa move when overall planet rotation is added into the calculation? Do the visible cloud tops here always (or just “usually”) correspond to a pressure altitude of 75 kPa? This particular movie synched planet rotation to the visual location of the Great Red Spot as Voyager arced in.

If one instead uses the value of G and Earth’s mass as determined by the University of Washington, Measurement of Newton’s Constant Using a Torsion Balance with Angular Acceleration Feedback (109 kb PDF, here), where Earth’s mass = 5.972245(82) × 10²⁴ kg and G = 6.674215(92) × 10^–11 m³kg^–1s^–2, then g = 9.82025(13) m/s². I used the U of W values when calculating the sea-level latitude (50.53°) at which g = 9.80665 m/s².

Regardless of which data set you use for your input variables, (producing values from 9.8226 using Wikipedia data or 9.82025(13) m/s² using U of W data), when one is comparing the gravity of the planets, g for Earth is not the latitude-based, centrifugally-adjusted value of 9.80665 m/s².

To make the table consistent across all planets, damn easy to calculate, and logical, I’ve revised it with surface gravities that are non-compensated for the centrifugal effects of some arbitrarily chosen latitude. If someone prefers a different solution, whatever it is, it must apply equally to all planets and must have the proper acceleration values for that latitude. If someone proposes to use equatorial accelerations, then all planets (including Earth, which has an equatorial value of 9.78033 m/s²) must be labeled with their equatorial values. Pluto could be footnoted that its value is an exception and is not an equatorial value. Really though, lowering one’s self down into Jupiter’s clouds to a pressure altitutde of 75 kPa (some old text I just deleted) and letting one’s self “rotate” with the planet (blow with the equatorial winds) is 1) hardly a “surface gravity”, and 2) is fraught with potential errors and the need for future recalculation because it is so dependent upon the methodology used in the measurment and calculation of equatorial winds, the altitude at which they are measured, and how one calculates the overall rotational period of a big gas ball. For instance, is “equatorial wind speed” measured relative to an inertial frame of reference(?), or to the planet-wide, mean cloud-top rotational period(?), or is the equatorial wind speed relative to the rotation rate of the planet’s magnetic core?

While editing the table, I also deleted the “formula” under it for calculating surface gravity. It had only two digits of precision and required the calculation of average planet density in kilograms per cubic meter. Beyond those shortcomings, it also had nothing to do with the old table because it had no terms to adjust for the rotational speed of the planet. I replaced that formula with the actual one for calculating gravity. If the value for G changes in the future, and/or the volume or mass of a planet is revised, any editor can easily and accurately update the new table. Better yet, any editor can audit the table and others’ calculations. Greg L (my talk) 18:59, 6 February 2008 (UTC)

Conversion of Meters to Feet

The conversion from meters to feet of Standard gravity

9.80665 m/s² (32.1740 ft/s²) appears an incorrect calculation.

9.80665 m = 382.45935 in = 31.8716125 ft

32.1740 ft = 386.088 in = 9.89969 m

A difference of 3.62865 in = 0.09304 m = 0.3023875 ft

Can someone explain the differences in calculation? Galo1969X (talk) 22:51, 16 February 2008 (UTC)

• It appears you are using a conversion factor of 39 inches per meter, which is clearly incorrect. The precise conversion from meters to feet is 0.3048 m/ft (12 inches × 2.54 cm/inch). Thus, there are 100 ÷ 2.54 ≈ 39.3700787401574803149606299213 inches/m. Thus, 9.80665 m/s² ÷ 0.3048 ≈ 32.1740485564304461942257217848 ft/s². Greg L (my talk) 05:25, 22 February 2008 (UTC)

Question

So an object falling to Earth from infinity would reach what velocity? —Preceding unsigned comment added by 216.56.28.38 (talk) 22:02, 11 May 2010 (UTC)

Wouldn't it be zero? If the object is an infinite distance away, then it would be out of the gravity well of the Earth and thus not subject to any acceleration. topace10 (talk) 04:43, 15 June 2010 (UTC)

The question implies that it would somehow start moving toward Earth by a teeny tiny speed (that would be subtracted out in the end). It also implies that it's not really infinity, just "very very far away". In which case, its speed when it reached the surface would be the same as the escape speed (escape velocity) from the surface, which is 11,200 m/s. 108.7.163.81 (talk) 06:07, 22 August 2010 (UTC)

In terms of spherical harmonics?

Putting it in terms of 'at Earth surface' seems rather arbitrary to me, and makes it harder to relate to what it is in space.88.159.72.164 (talk) 12:49, 28 May 2010 (UTC)

Are these gravities accurate for the given cities?

I just asked Wolfram Alpha (http://www.wolframalpha.com/) what the force of gravity was in Montréal and it gave me the number I added to the table (absolute magnitude); but when I computed it for Athens and Amsterdam, the numbers were a bit lower than those in the table. The Wolfram Alpha query gives the gravitational force in 3 dimensions, with a downward component as well as a southward and westward components for a full definition of the force experienced. However, the westward and southward components are exceedingly small, at about a 400,000th the force in the downward direction. In my opinion, I think we should adapt the Wolfram numbers as they seem better researched, but we should decide whether we want to depict the total magnitude, the downward component only, or all 4 values in their own columns which would make the table fill the horizontal space a bit better and give an interesting picture of the force in 3 dimensions.

--TimeHorse (talk) 14:44, 11 June 2010 (UTC)

Mr. TimeHorse, I am going to edit the Montreal number down to the same number of significant figures as the other cities. It looks rather odd as-is. —Preceding unsigned comment added by 144.5.140.24 (talk) 16:32, 15 June 2010 (UTC)

I have also just checked Wolfram Alpha and get a different number for Sydney, Australia (9.79607 total or 9.79602 downward) to what is given in the table (9.797). It's only a small difference, but I don't know whether the original tabulation contained a misprint, whether Wolfram Alpha's algorithm was still under development (on the way to Wolfram Beta?!), ...or whether gravity is gradually changing since someone looked last. The output I see says "based on EGM2008 12th order model; 37 meters [sic] above sea level". Perhaps a note like this should appear in the article/citation.

—DIV (137.111.13.4 (talk) 07:04, 26 August 2014 (UTC))

Proposal to move to Earth's gravity

A few months ago, this page was moved from Earth's gravity without explanation. Now the vast majority of links to this page are redirects from Earth's gravity. Aside from this, I don't see any strong reason for preferring one title over the other. --RockMagnetist (talk) 01:01, 24 October 2010 (UTC)

No real preference, though "Gravity of Earth" sounds more "encyclopedic" to me. The redirects are not much of a problem (quite a bunch of them are from anon talk page vandalism warnings...). Vsmith (talk) 01:23, 24 October 2010 (UTC)

"Gravity of the Earth" would be even more encyclopedic, being better English. --RockMagnetist (talk) 12:39, 24 October 2010 (UTC)

However, on the article page I see plenty of references to "Earth's gravity", which sounds right in context, while I can't picture saying "gravity of the Earth" in a sentence (I would say "the gravitational field of the Earth", or something similar.) --RockMagnetist (talk) 12:45, 24 October 2010 (UTC)

Those "Earth's gravity" instances within the article would be expected as that was the article title for most of its history. Question is: what is the most prevalent usage in published works? And was that the reason for the original title? I suspect it was, but don't know.

Looks like it was first created as Gravity (Earth) as a split from gravitation (section: The Earth's gravity) in March 2006 and moved to/renamed Earth's gravity on 20 June 2006, for a bit of article ancient history. Vsmith (talk) 00:57, 25 October 2010 (UTC)

I prefer the current title, "Gravity of Earth", to "Gravity of the Earth", which to me doesn't sound more professional or "encyclopedic". I believe that the term "the Earth" is an outdated and geocentric reference. It also creates difficulty in being consistent while recognizing other planets; for example, if there is (or were to be) a page describing Mars's gravity, such a page should be called "Gravity of Mars", not "Gravity of the Mars". - Johnlogic (talk) 21:40, 15 September 2011 (UTC)

Legends?

Shouldn't the charts have legends?

The rotating Geoid seems to have this legend. http://en.wikipedia.org/wiki/File:Geoids_sm.jpg

The image with the subsurface gravitational field strength needs an explanation, more detail, and a better legend. Web searches seem to bring up this article:

Gravitational field strength inside the Earth

Laurent Hodges

American Journal of Physics -- October 1991 -- Volume 59, Issue 10, pp. 954-

But, unfortunately it has no abstract, and the article isn't available for free on the web.--Keelec (talk) 10:09, 11 December 2010 (UTC)

Fallacy of the GRACE Project, Gravity from Space not the same as Gravity on surface of the planet.

Looking at the GRACE project Geoid, I quickly noticed that the highest gravity is at the mountain ranges (Andes, Rockys, Himalayans, etc.). This would be the gravity as seen from space which would include the highest mountain peaks (more rock), and parts of the planet closest to the satellite.

The problem is that the gravity experienced at the surface of the earth is also dependent on the distance from the center of the Earth. So, the gravity experienced at the surface of the earth from the highest peaks is actually LOWEST. In fact, in this article, it says that the gravity at the peak of Mt. Everest decreases by 0.28%, or 9.8 m/s2 --> 9.77 m/s2.

The Grace Geoid also doesn't seem to take into account the Centrifugal effects on gravity as experienced in the equatorial regions, or the effects of the "bulge" which effectively increases the altitude in the equatorial regions.--Keelec (talk) 03:50, 13 December 2010 (UTC)

=This guy seems to be claiming that the gravity felt in space above the Earth is not dependent on the distance to the center of the Earth, obviously a mistake. Its not clear to me, since he included the word "also" in an ambiguous way. The more rock you have underneath you, the higher the gravity, it seems to me.216.96.76.228 (talk) 01:23, 5 September 2013 (UTC)

No, obviously not a mistake but an interesting observation!

It seems to me that Keelec is clearly saying that a satellite-based observation will produce results for various lat-long combinations at a fixed (or almost fixed) distance from 'the' (or 'a selected') centre of the earth (so as to produce the "idealized smooth Earth"), whereas surface-based observation would also (!) be affected by measuring at different distances from that centre. From the numbers in the article and the Mt Everest example it looks like the surface effect would dominate the slab effect. I.e. and e.g.: if you would dig a vertical shaft (with negligible diameter) from the top of Mt Everest down to sea level then the gravity at the top of the shaft would be less than at its bottom, in spite of the additional slab. So a satellite-based observation of Mt Everest would indicate a larger than average g (adding the slab effect, but otherwise ignoring the height of Mt Everest), as opposed to the surface-based observation which (includes both effects and therefore) produces a lower value.

So this actually sounds like a perfectly sound and interesting observation to me. And that is without even addressing the issue of the centrifugal (pseudo)force and the bulge... AlexFekken (talk) 08:13, 30 December 2013 (UTC)

The Moon, not Luna

One or more unconfirmed users have been editing this page to change the Moon's name to Luna. The official name is the Moon - see for example NASA's page on the Moon. It is also the common usage: most people will be confused if you call it Luna. So, unless you can present some really strong arguments on this talk page, please stop messing around with the name. RockMagnetist (talk) 15:06, 16 October 2011 (UTC)

Value of g not the same as the reference.

Hey! Am I the only one to see that in the section "Mathematical models", the value of g = 9.780327( 1 + 0.0053024·sin.....etc, is NOT the same as the referenced text: This is the International Gravity Formula 1967, the 1967 Geodetic Reference System Formula, Helmert's equation or Clairaut's formula. http://geophysics.ou.edu/solid_earth/notes/potential/igf.htm

Where in there, the value g = 9.78031846 ( 1+ 0.0053024·sin.....etc, is used. Why this happens? — Preceding unsigned comment added by 85.74.145.68 (talk) 23:57, 2 May 2012 (UTC)

Plot in lead

A couple of times now, editors have decided that the legend for the plot in the lead (the gravity based on the GRACE mission) is wrong and have switched the roles of red and blue. Anyone else who is tempted to do this should look at the source for the image. The legend is correct! RockMagnetist (talk) 20:24, 1 June 2012 (UTC)

Perhaps an explanation of 'fallacy' (see my comments above) would help clear this up and prevent further changes. AlexFekken (talk) 08:53, 30 December 2013 (UTC)

@AlexFekken: That would be great. The problem is, I have yet to find a good source that interprets the global anomalies.RockMagnetist (talk) 19:11, 30 December 2013 (UTC)

Question about the constant gravitational acceleration?

Why all objects in free fall have the same gravitational acceleration “g” when diverse masses (say from lightest to heaviest) require unlike forces to move the same distance and can be inferred in the Newton’s 2nd law of motion which states “The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object” i.e. a= F/m 75.152.167.86 (talk) 06:56, 4 May 2013 (UTC)Eclectic Eccentric Kamikaze

Because the gravitational force is proportional to the mass of the object as well. A heavier object experiences a greater force. Reatlas (talk) 08:31, 4 May 2013 (UTC)

Why F1=ma=mg=F2 or a=g?

When a net force “F1” appeared in F1=ma generates a proportional acceleration while gravitational force of the same magnitude ”F2” of equation F2=GMm/r^2=mg produces constant gravitational acceleration if applied to a body of same mass “m”.

For example:

An acceleration of 4m/s/s is produced if a net 20N force “F1” is applied to block of mass of 5kg

F1=ma; a=F/m=20/5=4m/s/s.

A constant gravitational acceleration of g=9.8 m/s/s is NOT produced by the same magnitude of 20N gravitational force”F2” if acted on free fall mass of 5kg

F1 =ma=mg=F2=20=5x9.8 (Not dead-on)

Thus if two different types of acceleration are obtained by the same force (F1=F2=20N) if acted on the same mass (m=m=5kg) then why a=g? 75.152.167.86 (talk) 19:04, 8 May 2013 (UTC) Eclectic Eccentric Kamikaze

You can't randomly pick any value for F2 or gravitational force, the gravitational force equation is $F=G{\frac {m_{1}m_{2}}{r^{2}}}\$. G is gravitational constant=6.67*10^-11, m1 is mass Earth=5.97*10^24, m2 is 5kg and r is radius Earth=6,371,000. From this F2 is calculated to be 49.05N ≈ 5*9.81 = M*g Reatlas (talk) 02:17, 9 May 2013 (UTC)

Variation in Gravity by depth

The equation for the variation in gravity inside a sphere with linearly decreasing density is quoted as:

   $g(r)={\frac {4\pi }{3}}G\rho _{0}r-{\frac {4\pi }{3}}G\left(\rho _{0}-\rho _{1}\right){\frac {r^{2}}{r_{e}}}$ 

which is derived by substituting the density at radius r

   $\rho _{r}=\rho _{0}-\left(\rho _{0}-\rho _{1}\right){\frac {r}{r_{e}}}$

into the constant density equation:

   $g(r)={\frac {4\pi }{3}}GM={\frac {4\pi }{3}}G\rho r.$

However, this equation describes not the gravity at radius r for a sphere with varying density, but what the gravity at radius r would be if inside that radius the sphere were of uniform density $\rho _{r}$. These are not the same. If the density of the sphere enclosed at radius r varies linearly - from $\rho _{0}$ at the centre to $\rho _{r}$ at radius r (where $\rho _{0}>\rho _{r}$) - then the mass $M_{r}$ of the enclosed sphere should be the following integral:

   $\!M_{r}=\int _{0}^{r}4\pi \left(\rho _{0}-\left(\rho _{0}-\rho _{1}\right){\frac {r}{r_{e}}}\right)r^{2}dr$

Expanding this gives:

   $\!M_{r}=4\pi \rho _{0}\int _{0}^{r}r^{2}dr-4\pi {\frac {\left(\rho _{0}-\rho _{1}\right)}{r_{e}}}\int _{0}^{r}r^{3}dr$

Solving gives:

   $\!M_{r}=\left[{\frac {4}{3}}\pi \rho _{0}r^{3}\right]-\left[\pi {\frac {\left(\rho _{0}-\rho _{1}\right)}{r_{e}}}r^{4}\right]$

and hence by Gauss's Law

   $4\pi r^{2}g_{r}=4\pi GM_{r}$

$g(r)$ becomes:

   $g(r)=G\left[\left[{\frac {4}{3}}\pi \rho _{0}r\right]-\left[\pi {\frac {\left(\rho _{0}-\rho _{1}\right)}{r_{e}}}r^{2}\right]\right]$

   $g(r)={\frac {4}{3}}\pi G\rho _{0}r-\pi G\left(\rho _{0}-\rho _{1}\right){\frac {r^{2}}{r_{e}}}$

It therefore seems to me that the factor of ${\frac {4}{3}}$ should not appear in the second part of the equation as shown. Would someone who knows more maths and physics than I do please illuminate? George963 au (talk) 03:38, 6 May 2013 (UTC)

If the mass in a body is spherically symmetric, the gravity outside is the same as if all the mass were concentrated at the center (see shell theorem). It's a consequence of the inverse square law. So the gravity at the surface is the same for a constant density and a linearly increasing density, as long as the total mass is the same. I'll make this clear in the article as soon as I find a good reference. RockMagnetist (talk) 16:03, 6 May 2013 (UTC)

I agree that "the gravity at the surface is the same for a constant density and a linearly increasing density, as long as the total mass is the same". My query relates to the way the total mass of the sphere is being calculated from its linearly increasing/decreasing density. I think there is an error in the maths. Would you please double check? George963 au (talk) 02:45, 7 May 2013 (UTC)

Your math is correct and I have changed the equation in the article. Thanks for checking it! RockMagnetist (talk) 02:56, 7 May 2013 (UTC)

Thank you, RM, I am content. George963 au (talk) 03:15, 7 May 2013 (UTC)

It seems to me that there is some inconsistency in the sign convention in this section. Do people have any thoughts/comments on this? JCMPC (talk) 19:26, 12 April 2014 (UTC)

I checked the math again, and it still seems consistent to me. I take $\rho _{0}$ to be the density at the centre of the sphere; $\rho _{r}$ to be the density at radius r from the centre of the sphere; $\rho _{1}$ to be the density at the surface of the earth; and as the density decreases away from the centre, I take $\rho _{0}>\rho _{r}>\rho _{1}$. Where do you see an inconsistency? George963 au (talk) 14:08, 7 January 2015 (UTC)

Dubious introduction

"The gravity of Earth, denoted g, refers to the acceleration that the Earth imparts to objects on or near its surface. "

I think that this is wrong. Gravity is fundamentally a force existing between two objects which have mass. Not an acceleration. The force causes the apparent acceleration, not the other way around. The fact that some useful equations which describe this effect are conventionally formulated to include a constant (g) with the dimensionality of acceleration, does not alter this basic fact.Eregli bob (talk) 19:31, 2 August 2013 (UTC)

From a Newtonian 'force-as-action-at-a-distance' perspective you would be right. But I think that this is better and is meant to reflect the idea that gravity is just a pseudo-force with an acceleration independent of the mass of the object acted upon. The general theory of relativity is all about taking this to the next level, i.e. eliminating the force completely and replacing it with a curvature of space-time. So gravity becomes 'fundamentally' a curvature of space-time, the most noticeable aspect of which is a 'g'. AlexFekken (talk) 08:47, 30 December 2013 (UTC)

The Equatorial Bulge

I corrected a serious error. The bulge at the equator increases the gravitational force for the simple reason that more matter generates more gravity. The faulty account talks about the distance to the center, but the center is an imaginary point and doesn't weigh anything! — Preceding unsigned comment added by 74.83.182.22 (talk) 19:40, 2 August 2013 (UTC)

Does "g =GM/d^2" best in situ in F= GMm/d^2?

Does "g =GM/d²" which appear in Newton's universal law of gravitation conform with "F = GMm/d²" teleologically?

Where M= mass of earth, m= mass of an apple (smaller objects), d= on-center distance

It is inoculated that both M and m accelerate toward each other due to force of gravitation but m appears a lot to the M due to its greater acceleration as compared to the M toward m, which is so minuscule to be distinguished and withal, the reduction in "d" occurs due to falling of m (in its acceleration mode) ONLY, nonetheless, if the difference in M and m is not so colossal it stands to reason that both masses are changing their positions which conduce to decrease "d" due to the falling of both M and m (in the mode of their higher types of complex motion).

This can easily be observed if collated the following two identical spherical masses (from point to celestial) which are separated by on-center conspicuous distance “d”. Let

First Mass = M₁, Second Mass = M₂, M₁ = M₂ = Identical for simplicity, Centre-to-Centre distance b/w M₁ and M₂ = d, d₁ = d₂, d₁ + d₂ = d, Gravitational acceleration of M₁ = g₁, Gravitational acceleration of M₂ = g₂ and “c” be the mid point of “d".

According to the notion of Newton's universal law of gravitation, the precocious simultaneous procession of M₁ and M₂ in the forepart toward "c" is due to the coetaneous germination of g₁ and g₂, inter alia, the occurrence in reduction of "d" ("d₁" and "d₂" equally on both sides of “c”) and accretion in "F" are coeval which transform both M₁ and M₂ swiftly into higher derivatives of position w.r.t time such as gravitational jerk, jounce, crackle, pop, lock, drop etcetera.

Consequently, it is inferred presciently that two objects attract each other at much faster rate due to a posteriori development of the spawning of complex motion o'er time instead of advancing merely with accelerations (“g”) of "F" anecdotally. 162.157.234.1 (talk) 05:19, 27 January 2014 (UTC)Eclectic Eccentric Kamikaze

Equation issue

Surface gravitation - Somigliana's Equation: The existing equation claims being related to WGS84. The equation here http://mtp.mjmahoney.net/www/notes/altitude/altitude.html claims the same but has slightly different figures. Which one is now correct? — Preceding unsigned comment added by 81.214.27.247 (talk) 14:17, 19 June 2014 (UTC)

Cities

I suspect the figures for comparative cities may have changed slightly. For example the table here show Zurich at 9.807 but the source Wolfram Alpha gives 9.810 --Rumping (talk) 08:01, 19 November 2014 (UTC)

"Inertia" is not a suitable substitute for "Centrifugal (Pseudo)Force"

It's an unfortunate trend in WP and elsewhere to replace "Centrifugal Force" with "Inertia" because of the age-old misconfusion that CF isn't "real", etc. etc. etc. The trouble is that "inertia" is just one component of the situation, it doesn't adequately describe the phenomenon the whole world simply calls "centrifugal force". Saying "inertia" instead doesn't communicate anything. It would be ridiculous and off topic to be explaining why we're saying "inertia" all the time in contexts where people expect "centrifugal force". The idea of CF is perfectly intuitive, it's a bad idea to use language that's leads readers away from that intuition.
There are two better options:
1) If we just can't stand not being perfectly correct, we shouldn't make it worse by being even more wrong with "inertia", instead call it "centrifugal pseudoforce". "Centrifugal pseudoforce" means a pseudoforce pointing away from the center ("centrifugal" = "center fleeing"). That's exactly what it is.
or,
2) Just say "centrifugal force" and be done with it, and accept the fact that it's not 100% correct as trade for better communication. That way we don't have to explain "what is pseudoforce?" or "what's inertia referring to here?". The whole world knows what's meant by "centrifugal force". They don't know the arcane details, but the arcane details are off topic here.
I think option (2) is better because (1) suffers the same problem as "inertia" in that it needs to be explained.
100.0.124.147 (talk) 17:53, 3 March 2015 (UTC)

Well, I gave it six days and no one commented. Usually that means "tacit approval". I went ahead and clarified "effective gravity" vs. "gravity" in places where I saw it. I also rephrased the problematic "inertia" which was standing in for "centrifugal force" in a few spots. 100.0.124.147 (talk) 06:37, 9 March 2015 (UTC)

Your justification of making Wikipedia more accessible to lay readers by re-inserting flawed concepts like the centrifugal "force" is not satisfactory. The first and foremost criterion for scientific content on Wikipedia is that it must be accurate. Being accessible is a secondary consideration - many topics are not expected to be accessible to the lay reader because they require a lot of prior learning to be understood that the average reader is not expected to have. Nonetheless, we cover those areas and ensure that the content is accurate.

Some things need to be explained - this is what Wikipedia is for. As you've stated, there is no centrifugal "force". Whether one uses inertia or trajectory as the explanatory device is neither here nor there, but when explaining Earth's gravity, one cannot avoid discussing the difference between universal and geocentric frames of reference. The fact that we're discussing this shows very well that this needs to be explained.

Also, if you want to present things in a simpler fashion, there is the Simple English Wikipedia, which has an article on gravity that you might enjoy working on. Samsara 09:10, 9 March 2015 (UTC)

I actually don't mind it being explained as long as it's not muddled as it was before, and as long as it's concise because this is not actually an article about centrifugal pseudoforce (aka centrifugal force). Any explanation also needs to be not blitheringly incorrect. Simply calling it "centrifugal force" satisfied those requirements I thought. I proposed those solutions (1) and (2) for discussion, and in a void of response I had to go forward with a reasonable-enough fix. It certainly satisfies the requirements better than "inertia" (by itself) I think.
Hey, have you seen the famous XKCD cartoon regarding this? In the context of measuring net acceleration on the surface of the Earth, we are indeed using an accelerated reference frame, so the term does appear "plain as day" and is felt (well, measured). It surely is a real thing, it's just that its reality is maligned in the lay imagination by the misnomer "fictitious force", which is really only a bad alternative term for "pseudo force", "d'Alembert force", "inertial force" etc., which are all the same thing.
How about just calling it "centrifugal inertial force"? That's correct since "inertial force" == "pseudo force". And, it has the important descriptive term "centrifugal" in it, meaning "center fleeing". "Centrifugal" correctly cues readers that we're talking about the intuitive thing commonly called simply "centrifugal force". And, it's concise. It also might satisfy the buffs who like to reactively substitute "inertia" for "centrifugal force" because it has, well, the word "inertia" in it.
100.0.124.147 (talk) 10:28, 9 March 2015 (UTC)

We need to follow published work, creating neologisms is not acceptable. Samsara 10:34, 9 March 2015 (UTC)

Removed excess "centrifugal force" stuff. Vsmith (talk) 15:16, 9 March 2015 (UTC)

Samsara, that's a tall one. I think the burden of proof would be on you that "centrifugal force" is a neologism. :-) And, indeed we do need to follow published work, we need to follow the most reliable published work. Determining reliability is the responsibility of the editors using editorial knowledge. It's also our duty to communicate well. And the reactionary "inertia" in place of "centrifugal force" is poor communication because it makes a relatively clear meaning vague and confusing.
Follow-up: I just realized that I think you were referring to "centrifugal inertial force" as the neologism, and yes you have a good point there. Maybe if "centrifugal inertial force" were linked as so, the additional context would make it not a neologism? Just throwing ideas out there. :-)

Vsmith, before inserting "inertia" for "centrifugal force", I suggest you find reliable references which do that (use the single word "inertia" in place of centrifugal (pseudo)force). The trouble with it is that it's vague and incomplete for describing the idea of centrifugal force. You can find sources that do that, but they're uncommon and unreliable, and they also suffer from the same problem of being vague and incomplete for the job. You will find the most reliable and knowledgeable sources freely use "centrifugal" as a straightforward adjective meaning "away from the center". They are also clear as to what reference frame they're discussing (inertial vs. accelerating, etc.). They also assert strongly (and correctly) that in an accelerating reference frame such as on Earth's surface, the centrifugal "force" (as a shortened word for "pseudoforce") doesn't "not exist".

100.0.124.147 (talk) 19:13, 9 March 2015 (UTC)

My mistake. It seems that "centrifugal inertial force" has been used before. I'm not sure that you knew that when you proposed it. Samsara 21:51, 9 March 2015 (UTC)

You are quite right. My proposal for it's use was indeed based on the sense it made rather than on precedent, which I didn't know about. Of the four terms for the same thing (pseudo, d'Alembert, fictitious, and inertial) I personally prefer and tend to use "pseudo" for various reasons. But, I'm not hard-over about my choice of "pseudo" over the other words, just (as you know) about the misuse of "inertia" by itself as a substitute for "centrifugal force" (or for centrifugal pseudo/fictitious/dAlembert/inertial force). 100.0.124.147 (talk) 09:30, 10 March 2015 (UTC)

@100.0.124.147: Added <ref name="Boynton"> to first instance of "centrifugal force" which supports this term. Jim1138 (talk) 16:20, 10 March 2015 (UTC)

I can't tell you how much I appreciate that. It's a beautiful reference. 100.0.124.147 (talk) 23:29, 10 March 2015 (UTC)

Calculation of Fg Does Not Match Within Article

I noticed that the calculation of the force due to gravity does not match in the different sections (before their various mathematical adjustments) - $9.8331\,{\frac {\mathrm {m} }{\mathrm {s} ^{2}}}$ in Free Air Correction, and $9.822\,{\frac {\mathrm {m} }{\mathrm {s} ^{2}}}$ in Estimating g from the law of universal gravitation. Would anyone mind if I updated the values to match using a standard $G=6.673\ 84\times 10^{-11}{\rm {\ m^{3}\ kg^{-1}\ s^{-2}}}$ from Gravitational Constant? Goalie1998 (talk) 15:10, 26 May 2015 (UTC)

Acceleration?

How does gravity "accellerate" a stationary object on the surface of the Earth, as is stated in the first sentence of the article? 71.46.106.61 (talk) 00:24, 8 November 2015 (UTC)

The object is going around in a circle with the earth's surface instead of going in a straight line. That requires acceleration. RockMagnetist(talk) 06:28, 8 November 2015 (UTC)

Error in Free air correction section?

I am pretty sure the last equation in the Free air correction section is wrong. It certainly isn't consistent with the preceding equation. I think it should be g{φ,h} = g{φ} [1 − 3.155 × 10^-7 h]. Srw137 (talk) 21:15, 10 January 2016 (UTC)